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\begin{center}
	{\erhao \CJKfamily{kd}{2019年南开大学数学分析试题解答} }\\
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	{\sihao \textbf{满分：150分，考试时间：150分钟} }\\
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\begin{smybox}[colbacktitle=green!25!black!10!white]{第8题}
	已知$\alpha,\beta$均为正实数，且$\max\left\{\alpha ,\beta\right\}>1$，试证：
	\[
	\lim_{x\rightarrow +\infty}\int_1^x{\frac{1}{x^{\alpha}+t^{\beta}}}\mathrm{d}t=0
	\]
\begin{proof}由题设易知有
	\begin{enumerate} 
		\item 当$\alpha>1$时，有
		\[
		0 \leq \int_{1}^{x} \frac{\mathrm{d}t t}{x^{\alpha}+t^{\beta}} \leq \int_{1}^{x} \frac{1}{x^{\alpha}} \mathrm{d}tt=\frac{x-1}{x^{\alpha}} \rightarrow 0(x \rightarrow \infty)
		\]
		\item 当$\beta>1,0<\alpha \leq 1$时，且$\left(x\rightarrow\infty\right)$有
		\[
		\int_1^x{\frac{\mathrm{d}t}{x^{\alpha}+t^{\beta}}}=x^{-\alpha +\frac{\alpha}{\beta}}\int_1^x{\frac{\mathrm{d}\left(x^{-\frac{\alpha}{\beta}}t\right)}{1+\left(x^{-\frac{\alpha}{\beta}}t\right)^{\beta}}}\leq x^{-\alpha +\frac{a}{\beta}}\int_1^{1-\frac{\alpha}{\beta}}{\frac{\mathrm{d}u}{1+u^{\beta}}}\leq\frac{1}{\left(\beta -1\right)x^{\alpha\left(1-\frac{1}{\beta}\right)}}\rightarrow 0
		\]
	\end{enumerate}
\end{proof}
\end{smybox}
\begin{smybox}[colbacktitle=blue!75!black]{第9题}
	设$f(x)$在$[0,1]$上连续可微且恒不等于$0$，且$\int_0^1{f\left(x\right)}\mathrm{d}x=0$，证明：
	\[\int_0^1{\left| f\left(x\right)\right|}\mathrm{d}x\int_0^1{\left| f'\left(x\right)\right|\mathrm{d}x}>2\int_0^1{f^2\left(x\right)}\mathrm{d}x\]
\begin{proof}
	法1：
	\begin{align*}
	2\int_0^1{f}\left(x\right)^2\mathrm{d}x&=\int_0^1{f}\left(x\right)\left[2f\left(x\right)-f\left(0\right)-f\left(1\right)\right]\mathrm{d}x=\int_0^1{f}\left(x\right)\left[\left(f\left(x\right)-f\left(0\right)\right)-\left(f\left(1\right)-f\left(x\right)\right)\right]\mathrm{d}x\\
	&
	=\int_0^1{f}\left(x\right)\left[\int_0^x{f'}\left(t\right)\mathrm{d}t-\int_x^1{f'}\left(t\right)\mathrm{d}t\right]\mathrm{d}x\leq\int_0^1{\left| f\left(x\right)\right|}\left|\int_0^x{f'}\left(t\right)\mathrm{d}t-\int_x^1{f'}\left(t\right)\mathrm{d}t\right|\mathrm{d}x\\
	&
	\leq\int_0^1{\left| f\left(x\right)\right|}\left(\int_0^x{\left| f'\left(t\right)\right|}\mathrm{d}t+\int_x^1{\left| f'\left(t\right)\right|}\mathrm{d}t\right)\mathrm{d}x=\int_0^1{\left| f\left(x\right)\right|}\left(\int_0^1{\left| f'\left(t\right)\right|}\mathrm{d}t\right)\mathrm{d}x\\
	&=\int_0^1{\left| f'\left(x\right)\right|}\mathrm{d}x\cdot\int_0^1{\left| f\left(x\right)\right|}\mathrm{d}x
	\end{align*}
	法2：令$F\left(x\right)=\int_0^x{f}\left(t\right)\mathrm{d}t$，则有
	\[
	\int_0^1{f^2}\left(x\right)\mathrm{d}x=\left[f\left(x\right)F\left(x\right)\right]\mid_{0}^{1}-\int_0^1{f'}\left(x\right)F\left(x\right)\mathrm{d}x=-\int_0^{1}f'(x)F\left(x\right)\mathrm{d}x
	\]
	因此
	\[
	2\left| F\left(x\right)\right|=\left| F\left(x\right)-F\left(0\right)\right|+\left| F\left(1\right)-F\left(x\right)\right|=\left|\int_0^x{f}\left(t\right)\mathrm{d}t\right|+\left|\int_x^1{f}\left(t\right)\mathrm{d}t\right|\leq\int_0^1{\left| f\left(t\right)\right|}\mathrm{d}t
	\]
%	则有
%	\[\int_0^1{\left| f\left(x\right)\right|}\mathrm{d}x\int_0^1{\left| f'\left(x\right)\right|\mathrm{d}x}>2\int_0^1{f^2\left(x\right)}\mathrm{d}x\]
	%	注意到
	%	\begin{equation}
	%	\int_0^1{f^2}\left(x\right)\mathrm{d}x=\int_0^1{f\left(x\right)}\left(f\left(x\right)-f\left(0\right)\right)\mathrm{d}x=\int_0^1{f\left(x\right)}\left(\int_0^x{f'\left(t\right)}\mathrm{d}t\right)\mathrm{d}x
	%	\end{equation}
	%	\begin{equation}
	%	\int_0^1{f^2}\left(x\right)\mathrm{d}x=\int_0^1{f\left(x\right)\left(f\left(1\right)-f\left(x\right)\right)}\mathrm{d}x=\int_0^1{f\left(x\right)}\left(\int_x^1{f'\left(t\right)}\mathrm{d}t\right)\mathrm{d}x
	%	\end{equation}
	%	两式相加可得
	%	\[
	%	2\int_0^1{f^2}\left(x\right)\mathrm{d}x=\int_0^1{f\left(x\right)\left(\int_0^1{f'\left(x\right)}\mathrm{d}x\right)\mathrm{d}x}<\int_0^1{\left| f\left(x\right)\right|}\mathrm{d}x\int_0^1{\left| f'\left(x\right)\right|\mathrm{d}x}
	%	\]
	
%	法3：对$\int_{0}^{1}\left|f^{\prime}(x)\right| \mathrm{d} x \geq|f(b)-f(a)|, f(b)=\max _{x \in[0,1]} f(x), f(a)=\min _{x \in[0,1]} f(x)$，其中$a \in W=\{f(x)<0\}, b \in V=\{f(x) \geq 0\}$且$\int_{V} f(x) \mathrm{d} x+\int_{W} f(x) \mathrm{d} x=0$，即有
%	\begin{align*}
%	\int_0^1{\left| f'(x)\right|}\mathrm{d}x&		\geq\frac{\int_W{f^2}\left(x\right)\mathrm{d}x}{\int_W{f}\left(x\right)\mathrm{d}x}-\frac{\int_V{f^2}\left(x\right)\mathrm{d}x}{\int_V{f}\left(x\right)\mathrm{d}x}=\frac{\int_W{f^2}\left(x\right)\mathrm{d}x+\int_V{f^2}\left(x\right)\mathrm{d}x}{\int_W{f}\left(x\right)\mathrm{d}x}\\
%	&=\frac{\int_0^1{f^2}\left(x\right)\mathrm{d}x}{\int_W{f}\left(x\right)\mathrm{d}x}=\frac{\int_0^1{f^2}\left(x\right)\mathrm{d}x}{\frac{1}{2}\int_0^1{\left| f\left(x\right)\right|}\mathrm{d}x}
%	\end{align*}
	%		因此$\int_{0}^{1}\left|f^{\prime}(x)\right| \mathrm{d} x \int_{0}^{1}|f(x)| \mathrm{d} x \geq 2 \int_{0}^{1} f^{2}(x) \mathrm{d} x$
\end{proof}
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